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What is the solution to x+y=0?   

          x + y = 0   

               x  =  –y   

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Here is the processed version.

Given \(30-60-90\) triangle,

Hypotenuse \(=\) \(q\sqrt3\)

The short side that is opposite of the \(30\) degree angle is \(p\sqrt2\)

And the long side that is opposite to the \(60\) degree angle is \(p\sqrt6\)

Find the values of \(p\) and \(q.\)

Anyways, \(2p\sqrt2 = q\sqrt3\), techincally, there is 1 equation and 2 variables, making this problem more difficult to solve.

I'm going to have this problem solved by someone else (Because unfortunately, I must do my homework). 

Yeah, I managed to solve it 30 minutes later. (Sorry, the correct answer was -1.)

Anyways, thanks for the answer. What I did was take

\(z + \frac{1}{z}=1\) and multiply it by \(z^2\), getting \(z^3-z^2+z=0\). However, when you multiply \(z+ \frac{1}{z}=1\) by \(z\), you will get

\(z^2-z=1\)

From which you can replace the \(-z^2+z\) with \(-1\) (The negative of the equation above) to get

\(z^3-1=0\)

From which you can easily get \(z^3=-1\)

We are given a 30-60-90 triangle ABC with sides p6​, p2​, and q3​. We are asked to find the values of p and q. We can start by using the fact that the sides of a 30-60-90 triangle are in the ratio 1:3​:2. This means that we can write the following equations:

p2​p6​​=3​q3​p2​​=1

Simplifying these equations, we get:

2​6​​=3​3​2​​=q1​

Solving the first equation, we find that 6​=6​. This is true, so our first equation is valid.

Solving the second equation, we find that q=6​.

Therefore, the values of p and q are p = 1/sqrt(3) and q = sqrt(6)

Squaring the equation z + 1/z = 1 gives z^2 + 2 + 1/z^2 = 1. Then multiplying by z^2 gives z^4 + 2z^2 + 1 = z^2. Rearranging, we get z^4 + 2z^2 + 1 - z^2 = 0, which factors as (z^2 + 1)(z^2 + 1) = 0. Since z^2 + 1 = 0 has no solutions, z^2 = -1. Taking the square root of both sides, we find z = i or z = -i.

If z = i, then z^3 = i^3 = -i.

Alternatively, if z = -i, then z^3 is also equal to -i.

Therefore,  in either case, z^3 = -i.

What side is \(\frac{p}{\sqrt2}\)? What side is \(\frac{p}{\sqrt6}?\)

angle PCD = 75^o  

Grab a scrap of paper and sketch it. 

The line from P to C is the base of a isoceles triangle. 

Its equal sides are a side of the square and a side of the triangle.  

The big angle of the isoceles triangle is the sum of the 60^o of the  

equilateral triangle and the 90^o of the square, for a total of 150^o.   

The sum of the interior angles of any triangle is 180^o and so that   

leaves 30^o for the two small angles to share, namely 15^o each.  

At corner C, subtract that 15^o from the 90^o of the square, leaving 75^o.  

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What's this? ChatGPT spam in another language?

Da nacrtamo parabolu sa osom simetrije $x = 3$, $x$-presjekom na $x = -8$ i $y$-presjekom $5$, možemo početi identifikacijom jednačine parabole.

Jednačina parabole sa osom simetrije $x = a$ i $y$-presjekom $b$ data je na sljedeći način:

$$y = ax^2 + b$$

U ovom slučaju znamo da je osa simetrije $x = 3$, pa možemo zapisati jednačinu kao:

$$y = 3x^2 + 5$$

Sada možemo pronaći $x$-presret postavljanjem $y = 0$ i rješavanjem za $x$.

$$0 = 3x^2 + 5$$

Oduzimanje $5$ sa obe strane dobijamo:

$$-5 = 3x^2$$

Podijelimo obje strane sa $3$ daje nam:

$$-1/3 = x^2$$

Uzimanje kvadratnog korijena obje strane daje nam:

$$x = \pm \sqrt{1/3}$$

Dakle, $x$-presretak je $x = -2\sqrt{1/3} = -8$.

Sada možemo prikazati parabolu pomoću jednačine $y = 3x^2 + 5$. Da bismo to učinili, prvo možemo nacrtati $x$-presretak, koji je $x = -8$, a zatim pomoću jednačine pronaći odgovarajuće $y$-vrijednosti za svaku $x$-vrijednost.

Evo grafika parabole:

📈

Kao što vidite, parabola se otvara prema gore i ima minimalnu vrijednost od $5$ na $x = -8$. Osa simetrije je $x = 3$, a $x$-presjek je $x = -8$.

Da bismo pronašli površinu regije okružene grafikom od $x^2 + y^2 = 2x - 6y + 6 + 14x - 16y + 80$, možemo početi pojednostavljivanjem jednačine. Dopunivši kvadrat za termine $x$ i $y$, dobijamo: $x^2 ​​- 2x + y^2 + 6y = 86$ Sabiranjem i oduzimanjem konstanti za kompletiranje kvadrata imamo: $(x^2 - 2x + 1) + (y^2 + 6y + 9) = 86 + 1 + 9$ Ovo pojednostavljuje na: $(x - 1)^2 + (y + 3)^2 = 96$ Upoređujući ovo sa standardnim oblikom jednačine kružnice, $(x - h)^2 + (y - k)^2 = r^2$, možemo identifikovati da data jednadžba predstavlja krug sa centrom na $(1, -3)$ i poluprečnik $\sqrt{96}$. Površina kruga je data formulom $A = \pi r^2$. Zamjenom radijusa u ovu formulu dobijamo: $A = \pi \cdot 96$ Prema tome, površina regiona okruženog grafikom $x^2 + y^2 = 2x - 6y + 6 + 14x - 16y + 80$ iznosi $96\pi$ kvadratnih jedinica.

This should help (it uses the chain rule):

Since PZ bisects ∠QPR, ∠QPZ=∠RPZ. Also, since QZ=4 and PZ=2RP​=29​, we can find the length of PZ using the Pythagorean Theorem:

QZ2+PZ2=PR2.

Substituting the given values, we have

42+(29​)2=92.

Solving for PZ, we get PZ=215​.

Since ∠QPR is a right angle and ∠RPZ=∠QPZ, ∠RPZ and ∠QPZ are each complementary to ∠QPR. Therefore,

∠QPZ+∠RPZ=90∘.

Substituting the given values, we have

∠QPZ+∠RPZ=90∘,

so ∠QPZ=∠RPZ=290∘​=45∘.

Now, since ∠RPZ=∠QPZ=45∘, triangle RPZ is a 45-45-90 triangle, and RZ=15/sqrt(2).

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The axis of symmetry of the graph of y=ax2+bx+c is x=−2ab​. In general, the axis of symmetry is a vertical line, so we have −2ab​=−1. Solving for b, we find b=2a​=16.